Wednesday
June 19, 2013

# Posts by Margie

Total # Posts: 199

(2x-3)(x^2+5x-3)=2x+10x^2-15x+9 *Is this correct.? this is a different problem... Here's how this one works out: ......x^2 + 5x - 3 ............2x - 3 ================== ...... -3x^2 - 15x + 9 2x^3 + 10x^2 - 6x ====================== 2x^3 + 7x^2 - 21x + 9 I hope this will ...

3a^3b^2(-a^3b-a^2b^2)=-3a^9b^3-3a^6b^4 *is this correct.?- Remember to add exponents when multiplying with the same base. (3a^3b^2)(-a^3b - a^2b^2) = -3a^6b^3 - 3a^5b^4 I hope this will help.

math2(count iblis)
-y^3(2x^2y-4xy^2)=-2x^2y^4+4xy^6 *is this correct.? No, in the last term, the y should be to the fifth power (3+2)

math
sqrt3x^5y^14 To simplify: sqrt(3 x^5 y^14) The square root of x^4 is x^2. The square root of y^14 is y^7. Can you determine what to do now with those hints? I hope this will help.

math
-4x^2(3x^2-x+1)=-12x^4+4x^3-1x^2 *Is this correct.? The last term should be -4x^2

math
-4x^2(3x^2-x+1) Multiply it term by term, and I will, or someone will check you.

5y^8 - 125 *you have to factor completley 5y^8 - 125 = 5(y^8 - 25) = 5(y^4 + 5)(y^4 - 5)= 5(y^4 + 5)[y^2 + sqrt(5)][y^2 - sqrt(5)]= 5(y^4 + 5)[y^2 + sqrt(5)]* [y + 5^(1/4)][y - 5^(1/4)] good work, good explaination

Math 4 (Count Ibilis)
Solve by factoring: One of the solutions to the quadratic equation 2x^2 - 9x – 5 is -1/2. What is the other solution? 2x^2 - 9x – 5 = (2x + 1)(x - 5) So, it's x = 5.

I THOUGHT THIS WAS A HELP SITE!
noone is helping.Please help me. - get a life. YOu posted the question 15 min ago. We dont just sit here waiting for questions.

Math 3
Solve by factoring: One of the solutions to the quadratic equation x^2 - 13x + 30 is 3. What is the other solution? Is it:(x+3) If 3 is a solution, then the quadratic contains a factor (x-3). The other factor is thus (x-10), and it follows that the other root is x = 10. thanks...

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